.. The batteries shown in the circuit in Fig. E26.24 have negligibly small internal resistances. Find the current through (a) the30 Ω resistor; (b) the20 Ω resistor; (c) thebattery

F the circuit with suggested directions for the current in each segment as shOWn below.
For a good initial suggestion 10V, since the battery has the larger emf, it determines the direction of the current;
the direction of the current is from the — to + terminal through the battery and from the + to — terminal through the
circuit
And the current should split at the junction 0. to other segments of the circuit

$$\begin{gathered} {\mathrm{l}}_1=\frac{10}{30} \\ {\mathrm{l}}_1=0.333\mathrm{A} \\ 10.00-20{\mathrm{l}}_2-5\mathrm{V}=0 \end{gathered}$$

The current through 30 ohms resistor is 0.333A

(b)ApplyingKirchhoff’sloopruletothelargeloop,weget:
$$\begin{gathered} {\mathrm{I}}_2=\frac{10-5}{20} \\ {\mathrm{I}}_2=0.250\mathrm{A} \end{gathered}$$

(c) Finally, We apply Kirchhoff’s junction rule to the junction a, so We get:
$$\begin{gathered} \mathrm{I}-{\mathrm{I}}_1-{\mathrm{I}}_2=0 \\ \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2 \\ =0.333+0.250 \\ =0.583\mathrm{A} \\ \mathrm{I}=0.583\mathrm{A} \end{gathered}$$

Therefore the current through 20 ohms resistor is 0.250A and current through 10V battery is 0.583A